To prove how limit exists using definition given.
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The definition of a limit of a complex function goes as follows:- Let a function f be defined at all points z in some deleted neighbourhood of z°. Then the limit of f(z) as z approaches z° is a number w° . That means for each positive number ε , there is a positive number δ such that |f(z)- w°| is less than ε whenever| z-z° | lies between 0 & δ.
Now I am asked to prove
i). limiting value of ( Re z) as z tends to z° equals (Re z°).
ii). limiting value of conjugate of z equals conjugate of z° as z tends to z°.
using definition given above.
I have difficulty as I don't know how to start.
complex-analysis limits
edited Jan 30 '16 at 15:37
Martin Sleziak
43.5k6113260
asked Jan 30 '16 at 14:17
Kavita Sahu
13010
add a comment |Â
up vote
1
down vote
favorite
The definition of a limit of a complex function goes as follows:- Let a function f be defined at all points z in some deleted neighbourhood of z°. Then the limit of f(z) as z approaches z° is a number w° . That means for each positive number ε , there is a positive number δ such that |f(z)- w°| is less than ε whenever| z-z° | lies between 0 & δ.
Now I am asked to prove
i). limiting value of ( Re z) as z tends to z° equals (Re z°).
ii). limiting value of conjugate of z equals conjugate of z° as z tends to z°.
using definition given above.
I have difficulty as I don't know how to start.
complex-analysis limits
edited Jan 30 '16 at 15:37
Martin Sleziak
43.5k6113260
asked Jan 30 '16 at 14:17
Kavita Sahu
13010
Start by noting that $mathrmRe(z)-mathrmRe(z_0) = mathrmRe(z-z_0)$ and $overlinez-overlinez_0=overlinez-z_0$. Now try to show that $|mathrmRe(z-z_0)|$, $|overlinez-z_0|$ cannot be very large if $|z-z_0|$ is small.
– Jonathan Y.
Jan 30 '16 at 14:46
Start with $|Re(z)-Re(z^o)|=|Re(z-z^o)|leq |z-z^o|$ and $|Im(z)-Im(z^o)|=|Im(z-z^o)|le |z-z^o|.$
– DanielWainfleet
Jul 12 at 3:36
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
The definition of a limit of a complex function goes as follows:- Let a function f be defined at all points z in some deleted neighbourhood of z°. Then the limit of f(z) as z approaches z° is a number w° . That means for each positive number ε , there is a positive number δ such that |f(z)- w°| is less than ε whenever| z-z° | lies between 0 & δ.
Now I am asked to prove
i). limiting value of ( Re z) as z tends to z° equals (Re z°).
ii). limiting value of conjugate of z equals conjugate of z° as z tends to z°.
using definition given above.
I have difficulty as I don't know how to start.
complex-analysis limits
edited Jan 30 '16 at 15:37
Martin Sleziak
43.5k6113260
asked Jan 30 '16 at 14:17
Kavita Sahu
13010
The definition of a limit of a complex function goes as follows:- Let a function f be defined at all points z in some deleted neighbourhood of z°. Then the limit of f(z) as z approaches z° is a number w° . That means for each positive number ε , there is a positive number δ such that |f(z)- w°| is less than ε whenever| z-z° | lies between 0 & δ.
Now I am asked to prove
i). limiting value of ( Re z) as z tends to z° equals (Re z°).
ii). limiting value of conjugate of z equals conjugate of z° as z tends to z°.
using definition given above.
I have difficulty as I don't know how to start.
complex-analysis limits
edited Jan 30 '16 at 15:37
Martin Sleziak
43.5k6113260
asked Jan 30 '16 at 14:17
Kavita Sahu
13010
edited Jan 30 '16 at 15:37
Martin Sleziak
43.5k6113260
edited Jan 30 '16 at 15:37
Martin Sleziak
43.5k6113260
edited Jan 30 '16 at 15:37
Martin Sleziak
43.5k6113260
43.5k6113260
asked Jan 30 '16 at 14:17
Kavita Sahu
13010
asked Jan 30 '16 at 14:17
Kavita Sahu
13010
asked Jan 30 '16 at 14:17
Kavita Sahu
13010
13010
Start by noting that $mathrmRe(z)-mathrmRe(z_0) = mathrmRe(z-z_0)$ and $overlinez-overlinez_0=overlinez-z_0$. Now try to show that $|mathrmRe(z-z_0)|$, $|overlinez-z_0|$ cannot be very large if $|z-z_0|$ is small.
– Jonathan Y.
Jan 30 '16 at 14:46
Start with $|Re(z)-Re(z^o)|=|Re(z-z^o)|leq |z-z^o|$ and $|Im(z)-Im(z^o)|=|Im(z-z^o)|le |z-z^o|.$
– DanielWainfleet
Jul 12 at 3:36
add a comment |Â
Start by noting that $mathrmRe(z)-mathrmRe(z_0) = mathrmRe(z-z_0)$ and $overlinez-overlinez_0=overlinez-z_0$. Now try to show that $|mathrmRe(z-z_0)|$, $|overlinez-z_0|$ cannot be very large if $|z-z_0|$ is small.
– Jonathan Y.
Jan 30 '16 at 14:46
Start with $|Re(z)-Re(z^o)|=|Re(z-z^o)|leq |z-z^o|$ and $|Im(z)-Im(z^o)|=|Im(z-z^o)|le |z-z^o|.$
– DanielWainfleet
Jul 12 at 3:36
Start by noting that $mathrmRe(z)-mathrmRe(z_0) = mathrmRe(z-z_0)$ and $overlinez-overlinez_0=overlinez-z_0$. Now try to show that $|mathrmRe(z-z_0)|$, $|overlinez-z_0|$ cannot be very large if $|z-z_0|$ is small.
– Jonathan Y.
Jan 30 '16 at 14:46
Start by noting that $mathrmRe(z)-mathrmRe(z_0) = mathrmRe(z-z_0)$ and $overlinez-overlinez_0=overlinez-z_0$. Now try to show that $|mathrmRe(z-z_0)|$, $|overlinez-z_0|$ cannot be very large if $|z-z_0|$ is small.
– Jonathan Y.
Jan 30 '16 at 14:46
Start with $|Re(z)-Re(z^o)|=|Re(z-z^o)|leq |z-z^o|$ and $|Im(z)-Im(z^o)|=|Im(z-z^o)|le |z-z^o|.$
– DanielWainfleet
Jul 12 at 3:36
Start with $|Re(z)-Re(z^o)|=|Re(z-z^o)|leq |z-z^o|$ and $|Im(z)-Im(z^o)|=|Im(z-z^o)|le |z-z^o|.$
– DanielWainfleet
Jul 12 at 3:36
add a comment |Â
1 Answer
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i) Note that as $z to z_0$, $forall epsilon > 0,$ take $delta := epsilon$ such that $forall z: 0 < |z-z_0| < delta, $
$$|mathrmRe(z) - mathrmRe(z_0)| le sqrt(mathrmRe(z) - mathrmRe(z_0))^2 + (mathrmIm(z) - mathrmIm(z_0))^2 = |z-z_0|<epsilon.$$
$$textHence, lim_zto z_0 mathrmRe(z) = mathrmRe(z_0).$$
ii) As $z to z_0$, $forall epsilon > 0,$ take $delta := epsilon$ such that $forall z: 0 < |z-z_0| < delta, $
$$|overlinez-overlinez_0| = |overlinez-z_0| = |z-z_0|<epsilon.$$
$$textHence, lim_zto z_0 overlinez = overlinez_0.$$
answered Jan 30 '16 at 14:51
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
i) Note that as $z to z_0$, $forall epsilon > 0,$ take $delta := epsilon$ such that $forall z: 0 < |z-z_0| < delta, $
$$|mathrmRe(z) - mathrmRe(z_0)| le sqrt(mathrmRe(z) - mathrmRe(z_0))^2 + (mathrmIm(z) - mathrmIm(z_0))^2 = |z-z_0|<epsilon.$$
$$textHence, lim_zto z_0 mathrmRe(z) = mathrmRe(z_0).$$
ii) As $z to z_0$, $forall epsilon > 0,$ take $delta := epsilon$ such that $forall z: 0 < |z-z_0| < delta, $
$$|overlinez-overlinez_0| = |overlinez-z_0| = |z-z_0|<epsilon.$$
$$textHence, lim_zto z_0 overlinez = overlinez_0.$$
answered Jan 30 '16 at 14:51
GNU Supporter
11.8k72143
add a comment |Â
up vote
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i) Note that as $z to z_0$, $forall epsilon > 0,$ take $delta := epsilon$ such that $forall z: 0 < |z-z_0| < delta, $
$$|mathrmRe(z) - mathrmRe(z_0)| le sqrt(mathrmRe(z) - mathrmRe(z_0))^2 + (mathrmIm(z) - mathrmIm(z_0))^2 = |z-z_0|<epsilon.$$
$$textHence, lim_zto z_0 mathrmRe(z) = mathrmRe(z_0).$$
ii) As $z to z_0$, $forall epsilon > 0,$ take $delta := epsilon$ such that $forall z: 0 < |z-z_0| < delta, $
$$|overlinez-overlinez_0| = |overlinez-z_0| = |z-z_0|<epsilon.$$
$$textHence, lim_zto z_0 overlinez = overlinez_0.$$
answered Jan 30 '16 at 14:51
GNU Supporter
11.8k72143
add a comment |Â
up vote
0
down vote
up vote
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down vote
i) Note that as $z to z_0$, $forall epsilon > 0,$ take $delta := epsilon$ such that $forall z: 0 < |z-z_0| < delta, $
$$|mathrmRe(z) - mathrmRe(z_0)| le sqrt(mathrmRe(z) - mathrmRe(z_0))^2 + (mathrmIm(z) - mathrmIm(z_0))^2 = |z-z_0|<epsilon.$$
$$textHence, lim_zto z_0 mathrmRe(z) = mathrmRe(z_0).$$
ii) As $z to z_0$, $forall epsilon > 0,$ take $delta := epsilon$ such that $forall z: 0 < |z-z_0| < delta, $
$$|overlinez-overlinez_0| = |overlinez-z_0| = |z-z_0|<epsilon.$$
$$textHence, lim_zto z_0 overlinez = overlinez_0.$$
answered Jan 30 '16 at 14:51
GNU Supporter
11.8k72143
i) Note that as $z to z_0$, $forall epsilon > 0,$ take $delta := epsilon$ such that $forall z: 0 < |z-z_0| < delta, $
$$|mathrmRe(z) - mathrmRe(z_0)| le sqrt(mathrmRe(z) - mathrmRe(z_0))^2 + (mathrmIm(z) - mathrmIm(z_0))^2 = |z-z_0|<epsilon.$$
$$textHence, lim_zto z_0 mathrmRe(z) = mathrmRe(z_0).$$
ii) As $z to z_0$, $forall epsilon > 0,$ take $delta := epsilon$ such that $forall z: 0 < |z-z_0| < delta, $
$$|overlinez-overlinez_0| = |overlinez-z_0| = |z-z_0|<epsilon.$$
$$textHence, lim_zto z_0 overlinez = overlinez_0.$$
answered Jan 30 '16 at 14:51
GNU Supporter
11.8k72143
answered Jan 30 '16 at 14:51
GNU Supporter
11.8k72143
answered Jan 30 '16 at 14:51
GNU Supporter
11.8k72143
answered Jan 30 '16 at 14:51
GNU Supporter
11.8k72143
11.8k72143
add a comment |Â
add a comment |Â
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Start by noting that $mathrmRe(z)-mathrmRe(z_0) = mathrmRe(z-z_0)$ and $overlinez-overlinez_0=overlinez-z_0$. Now try to show that $|mathrmRe(z-z_0)|$, $|overlinez-z_0|$ cannot be very large if $|z-z_0|$ is small.
– Jonathan Y.
Jan 30 '16 at 14:46
Start with $|Re(z)-Re(z^o)|=|Re(z-z^o)|leq |z-z^o|$ and $|Im(z)-Im(z^o)|=|Im(z-z^o)|le |z-z^o|.$
– DanielWainfleet
Jul 12 at 3:36